The maximum value of `absz` when z satisfies the condition `abs(z+2/z)`=2 is :

Options

A. `sqrt3-1`
B. `sqrt3`
C. `sqrt3+1`
D. 1

Answer

To find the maximum value of \( |z| \) when \( |z + \frac{2}{z}| = 2 \), we can follow these steps:

### Step 1: Rewrite the given equation
We start with the equation:
\[ |z + \frac{2}{z}| = 2 \]

### Step 2: Apply the triangle inequality
Using the triangle inequality, we can express this as:
\[ |z| + |\frac{2}{z}| \geq |z + \frac{2}{z}| \]
Thus, we have:
\[ |z| + \frac{2}{|z|} \geq 2 \]

### Step 3: Rearrange the inequality
Rearranging gives us:
\[ |z| + \frac{2}{|z|} - 2 \geq 0 \]

### Step 4: Multiply through by |z|
Let \( |z| = r \). Then the inequality becomes:
\[ r + \frac{2}{r} - 2 \geq 0 \]
Multiplying through by \( r \) (assuming \( r > 0 \)):
\[ r^2 - 2r + 2 \geq 0 \]

### Step 5: Solve the quadratic inequality
To solve \( r^2 - 2r + 2 \geq 0 \), we find the roots of the equation:
\[ r^2 - 2r + 2 = 0 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \]
Since the roots are complex, the quadratic does not cross the x-axis, indicating it is always positive.

### Step 6: Find the maximum value of |z|
Now, we need to maximize \( r + \frac{2}{r} \) under the constraint \( r + \frac{2}{r} = 2 \). To find the maximum, we can use calculus or analyze the function:
\[ f(r) = r + \frac{2}{r} \]
Taking the derivative:
\[ f'(r) = 1 - \frac{2}{r^2} \]
Setting \( f'(r) = 0 \):
\[ 1 - \frac{2}{r^2} = 0 \implies r^2 = 2 \implies r = \sqrt{2} \]

### Step 7: Evaluate the function at critical points
Now we evaluate \( f(r) \) at \( r = \sqrt{2} \):
\[ f(\sqrt{2}) = \sqrt{2} + \frac{2}{\sqrt{2}} = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

### Step 8: Conclusion
Thus, the maximum value of \( |z| \) is:
\[ \max |z| = 1 + \sqrt{3} \]